What is the slope of the graph of the equation at the given point?

Question

1. x^2y + y^3x = -6 Point: (2,-1)
2. (4x + 4y)^3 = 64x^3 + 64y^3 Point: (-1,1)

*Please have answers for both number 1 and 2. They are separate questions but with the same concept.

Answer 1

x²y + y³x = -6

2xy + x²(dy/dx) + 3y²(dy/dx)x + y³ = 0

dy/dx(x² + 3y²) = -2xy - y³

dy/dx = (-2xy - y³) / (x² + 3y²)

So, at the point (2,-1), dy/dx = 5/7.

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Simplifying first, we have 64(x + y)³ = 64x³ + 64y³

From the Binomial Theorem, 64(x³ + 3x²y + 3xy² + y³) = 64x³ + 64y³

So, 192x²y + 192xy² = 0

Divide by 192 to obtain x²y + xy² = 0

Differentiating implicitly with respect to x, we obtain:

2xy + x²(dy/dx) + y² + 2xy(dy/dx) = 0

dy/dx(x² + 2xy) = -2xy - y²

So, dy/dx = (-2xy - y²) / (x² + 2xy)

At the point (-1,1), dy/dx = 1/(-1) = -1